Question: Complete the square to solve for $x$. $2x^{2}+x-1 = 0$
Explanation: First, divide the polynomial by $2$ , the coefficient of the $x^2$ term. $x^2 + \dfrac{1}{2}x - \dfrac{1}{2} = 0$ Move the constant term to the right side of the equation. $x^2 + \dfrac{1}{2}x = \dfrac{1}{2}$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $\dfrac{1}{2}$ , so half of it would be $\dfrac{1}{4}$ , and squaring it gives us ${\dfrac{1}{16}}$ $x^2 + \dfrac{1}{2}x { + \dfrac{1}{16}} = \dfrac{1}{2} { + \dfrac{1}{16}}$ We can now rewrite the left side of the equation as a squared term. $( x + \dfrac{1}{4} )^2 = \dfrac{9}{16}$ Take the square root of both sides. $x + \dfrac{1}{4} = \pm\dfrac{3}{4}$ Isolate $x$ to find the solution(s). $x = -\dfrac{1}{4}\pm\dfrac{3}{4}$ The solutions are: $x = \dfrac{1}{2} \text{ or } x = -1$ We already found the completed square: $( x + \dfrac{1}{4} )^2 = \dfrac{9}{16}$